Sociopath Community

3.5 The Derivative of the Function y = xⁿ

Theorem. The derivative of the function y = xⁿ, where n is a positive integer, is equal to nxⁿ⁻¹.

Proof.

We have a function y = xⁿ

If x receives an increment Δx, then

y + Δy = (x + ΔX)ⁿ

Solving for Δy,

Δy = (x + Δx)ⁿ - y = (x + Δx)ⁿ - xⁿ

Applying Newton's binomial theorem we find,

(x + Δx)ⁿ - xⁿ = xⁿ+ nxⁿ⁻¹Δx + (n(n-1)/2)xⁿ⁻²(Δx)² + . . . + (Δx)ⁿ - xⁿ

                    = nxⁿ⁻¹Δx + (n(n-1)/2)xⁿ⁻²(Δx)² + . . . + (Δx)ⁿ

We find the ratio,

Δy      nxⁿ⁻¹Δx + (n(n-1)/2)xⁿ⁻²Δx + . . . + (Δx)ⁿ
---- = --------------------------------------------------
Δx                              Δx

           Δx(nxⁿ⁻¹ + (n(n-1)/2)xⁿ⁻²Δx + . . . + (Δx)ⁿ⁻¹
       = -----------------------------------------------------
                                       Δx
       = nxⁿ⁻¹ + (n(n-1)/2)xⁿ⁻²Δx + . . . + (Δx)ⁿ⁻¹

The we find the limit of this ratio,

                  Δy
lim(Δx->0) ---- = lim(Δx->0)[nxⁿ⁻¹ + (n(n-1)/2)xⁿ⁻²Δx + . . . + (Δx)ⁿ⁻¹]
                  Δx
                        = nxⁿ⁻¹ + (n(n-1)/2)xⁿ⁻²(0) + . . . + (0)ⁿ⁻¹

                        = nxⁿ⁻¹

Therefore, y' = nxⁿ⁻¹

-----------------------------------------------------------
Examples : :
-----------------------------------------------------------

(1) y = x⁵
y' = 5x⁵⁻¹ = 5x⁴

(2) y = x
y'= 1x¹⁻¹ = x⁰ = 1

(3) y = Sqrt(x) = x½
y' = (1/2)x(½-1) = (1/2)x⁻½ = 1/(2Sqrt(x))

What a Mathematician and an Artist have in common?

They will both probably starve to death.

The Waltz said: 

What a Mathematician and an Artist have in common?

They will both probably starve to death.

 And humanity would lack great depth and beauty in the absence of either.

3.6 Derivatives of y=sinx and y=cosx

-----------------------------------------------------------
y = sinx : :
-----------------------------------------------------------

If y = sinx, then y' = cosx

Proof.

Increase the argument x by the increment Δx

y + Δy = sin(x + Δx)

Solve for Δy

Δy = sin(x + Δx) - y = sin(x + Δx) - sin(x)

By the identity sinα - sinβ = 2cos((α + β)/2)sin((α - Β)/2),

sin(x + Δx) - sin(x) = 2cos((x + Δx + x)/2)sin((x + Δx - x)/2)
                             = 2cos((2x + Δx)/2)sin(Δx/2)
                             = 2cos(x + Δx/2)sin(Δx/2)

Take the ratio of Δy and Δx,

Δy      2sin(Δx/2)cos(x + Δx/2)       2sin(Δx/2) cos(x + Δx/2)
---- = ------------------------------ =  ---------------
Δx                     Δx                           2Δx/2

                                                    sin(Δx/2) cos(x + Δx/2)
                                                = -------------
                                                        Δx/2

Take the limit of the ratio Δy/Δx as Δx -> 0,
               sin(Δx/2) cos(x + Δx/2)                      sin(Δx/2)
lim(Δx->) -----------                        = lim(Δx->) ----------- lim(Δx->0) cos(x + Δx/2)
                 Δx/2                                                   Δx/2

It is the case that,

                 sin(Δx/2)
lim(Δx->0) ----------- = 1
                     Δx/2

As such,

lim(Δx->0) cos(x + Δx/2) = cos(x + 0/2) = cosx

Therefore, y' = cosx

-----------------------------------------------------------
y = cosx : :
-----------------------------------------------------------

If y = cosx, then y' = -sinx

Proof.

Increase the argument x by the increment Δx,

y + Δy = cos(x + Δx)

Solve for Δy,

Δy = cos(x + Δx) - y = cos(x + Δx) - cos(x)

By the identity cosα - cosβ = -2sin((α + β)/2)sin((α - β)/2),

cos(x + Δx) - cos(x) = -2sin((x + Δx + x)/2)sin((x + Δx - x)/2)
                              = -2sin((2x + Δx)/2)sin(Δx/2)
                              = -2sin(x + Δx/2)sin(Δx/2)

Take the ratio of Δy and Δx,

Δy     -2sin(x + Δx/2)sin(Δx/2)     -sin(x + Δx/2) 2sin(Δx/2)
---- = ------------------------------ = ------------------
Δx                     Δx                            2Δx/2

                                                     -sin(x + Δx/2) sin(Δx/2)
                                                 =  -----------------
                                                             Δx/2

Take the limit of the ratio Δy/Δx as Δx -> 0,

                   Δy                       -sin(Δx/2)   sin(x + Δx/2)
lim(Δx->0) ------ = lim(Δx->0) -------------
                   Δx                             Δx/2

                                                sin(Δx/2)
                           = -lim(Δx->) ------------- lim(Δx->0)sin(x + Δx/2)
                                                   Δx/2

It is the case that,

                   sin(Δx/2)
-lim(Δx->0) ------------ = -1
                      Δx/2

As such,

-lim(Δx->0)sin(x + Δx/2) = -sin(x + 0/2) = -sin(x)

Therefore, y' = -sinx

"Derivatives and Integrals" Where are the integrals?

LiYang said: 

"Derivatives and Integrals" Where are the integrals?

 All in good time.

3.7 Derivatives of Constants, Sums, Products, and Quotients

-----------------------------------------------------------
Constant : :
-----------------------------------------------------------

If y = C, where C is constant, then y' = 0.

Proof.

Let y = C be a function of x such that the values of it are equal to C for all x.

hence, for any value of x,

y = f(x) = C

We increase the argument x by the increment Δx,

y + Δy = f(x + Δx)

Given the function y retains the value C for all values,

y + Δy = C

We solve for Δy,

Δy = C - y

Again, y retains the value C,

Δy = C - C = 0

Take the ratio of the increment Δy and Δx,

 Δy      0
---- = ---- = 0
 Δx     Δx

Take the limit of the ratio,

                 Δy
lim(Δx->0)---- = lim(Δx->0)0 = 0
                 Δx

Therefore, y' = 0

Ex. y = 5

y ' = 0

-----------------------------------------------------------
Constant Factor : :
-----------------------------------------------------------

If y = Cu(x), where C is constant, then y' = Cu'(x)

Proof.

Let y = Cu(x) where C is constant

Increment the value of x by Δx,

y + Δy = Cu(x + Δx)

Solve for Δy,

Δy = Cu(x + Δx) - y = Cu(x + Δx) - Cu(x)

Take the ratio of the increments Δy and Δx,

Δy      Cu(x + Δx) - Cu(x)     C(u(x + Δx) - u(x))
---- = ----------------------- = ----------------------
 Δx                Δx                           Δx

Take the limit of the ratio as Δx approaches 0,

                 Δy                     C(u(x + Δx) - u(x))
lim(Δx->0)---- = lim(Δx->0)----------------------
                 Δx                               Δx
                                          

                                             u(x + Δx) - u(x)
                       = Clim(Δx->0)--------------------
                                                       Δx

= Cu'(x)

therefore, y' = Cu'(x)

Ex. y = 3(1/Sqrt(x)) = 3(x-½)
y' = 3(1/2)x^(-1/2 - 1) = (3/2)x^(-3/2)

-----------------------------------------------------------
Sums : :
-----------------------------------------------------------

If y = u(x) + v(x) + w(x), then y' = u'(x) + v'(x) + w'(x)

Proof.

For the values of the argument x let y = u + v + w

We increment the argument x by Δx,

y + Δy = (u + Δu) + (v + Δv) + (w + Δw)

Solve for Δy,

Δy = (u + Δu) + (v + Δv) + (w + Δw) - y
     = (u + Δu) + (v + Δv) + (w + Δw) - (u + v + w)
     = u + Δu + v + Δv + w + Δw - u - v - w
     = Δu + Δv + Δw

Take the ratio of the increment Δy and Δx,

Δy      Δu + Δv + Δw     Δu     Δv    Δw
---- = ------------------ = ---- + ---- + ----
Δx               Δx              Δx     Δx     Δx

Take the limit of the ratio as Δx approaches 0,
                 Δy                      Δu     Δv     Δw
lim(Δx->0)---- = lim(Δx->0)[---- + ---- + ----]
                  Δx                      Δx     Δx     Δx

                                            Δu                     Δv                     Δw
                        = lim(Δx->0)---- + lim(Δx->0)---- + lim(Δx->0)----
                                            Δx                      Δx                     Δx 

                         = u'(x) + v'(x) + w'(x)

therefore, y' = u'(x) + v'(x) + w'(x)

Ex. y = 3x⁴ - 1/cubeRoot(x)
y' = (3*4)x³ - (-1/3)x^(-1/3) = 12x³ + (1/3)(1/xcubeRoot(x))

-----------------------------------------------------------
Products : :
-----------------------------------------------------------

If y = uv, then y' = u'v + uv'

Proof.

For values of the argument x let y = uv

Increment the argument x by Δx,

y + Δy = (u + Δu)(v + Δv)

Solve for Δy,

Δy = (u + Δu)(v + Δv) - uv
     = (uv + uΔv + vΔu + ΔuΔv - uv
     = uΔv + vΔu + ΔuΔv

Take the ratio of the increments Δy and Δx,

Δy      uΔv + vΔu + ΔuΔv      uΔv    vΔu    ΔuΔv
---- = ------------------------- = ----- + ----- + ------
Δx                  Δx                     Δx     Δx       Δx

Take the limit of the ratio as Δx approaches 0,

                 Δy                     uΔv     vΔu    ΔuΔv
lim(Δx->0)---- = lim(Δx->0) ----- + ----- + ------
                 Δx                        Δx      Δx       Δx

                                            Δv                      Δu                                          Δv
                      = ulim(Δx->0)---- +vlim(Δx->0)---- + lim(Δx->0)Δulim(Δx->0)----
                                            Δx                       Δx                                          Δx

                      = uv' + vu' + 0v'

                      = uv' vu'

therefore, y' = u'v + yv'

Ex. y = x²sinx

y' = 2xsinx + x²cosx

-----------------------------------------------------------
Quotients : :
-----------------------------------------------------------

          u                   u'v - uv'
If y = --- , then y' = ----------
          v                        v²

Proof.

For the values of the argument x let y = u/v

Increment the argument x by Δx,

              u + Δu
y + Δy = --------
               v + Δv

Solve for Δy,

        u + Δu         u + Δu    u
Δy = -------- - y = --------- - ---
        v + Δv          v + Δv     v

                           vΔu - uΔv
                        = -------------
                             v(v + Δv)

Take the ratio of the increments Δy and Δx,

         vΔu - uΔv      vΔu - uΔv
         -------------     -------------
Δy      v(v + Δv)             Δx
---- = -------------- = ------------
Δx           Δx            v(v + Δv)

                               vΔu    uΔv
                                ----- - -----
                                 Δx      Δx
                            = --------------
                                  v(v + Δv)

Take the limit of the ratio when Δx approaches 0,

                                           vΔu   uΔv
                                           ----- - -----
                Δy                         Δx     Δx
lim(Δx->0)---- = lim(Δx->0) --------------
                 Δx                        v(v + Δv)

                                            Δu                      Δv
                       = vlim(Δx->0)---- - ulim(Δx->0)----
                                             Δx                       Δx
                         ------------------------------------------
                                     vlim(Δx->0)[v + Δv]

                            vu' - uv'
                        = -----------
                                vv

                        vu' - uv'
therefore, y' = ----------
                              v²

Ex. y = x³/cosx

y' = (3xcosx + x³sinx)/cos²x

3.8 The Derivative of a Logarithmic Function

If y = logₐx, then y' = (1/x)logₐe

Proof.

For the argument x let y = logₐx

Increment the argument x by Δx,

y + Δy = logₐ(x + Δx)

Solve for Δy,

Δy = logₐ(x + Δx) - y = logₐ(x + Δx) - logₐx
                               = logₐ([x + Δx]/x)
                               = logₐ(1 + Δx/x)

Take the ratio of the increments Δy and Δx,

Δy      logₐ(1 + Δx/x)
---- = ------------------
Δx              Δx

          1
      = --- logₐ(1 + Δx/x)
         Δx

          1   x
      = --- --- logₐ(1 + Δx/x)
          x  Δx

          1
      = --- logₐ(1 + Δx/x)^(x/Δx)
          x

let the quantity Δx/x be denoted as α, and given this quantity stores Δx, α -> 0 as Δx -> 0,

Δy      1
---- = --- logₐ(1 + α)^(1/α)
Δx      x

It is given that lim(α->0)(1 + α)^(1/α) = e

Take the limit of the ratio,

                 Δy                      1                               1
lim(Δx->0)---- = lim(Δx->0)---logₐ(1 + α)^(1/α) = ---logₐe
                 Δx                      x                               x